Question: Let $h(x)=\begin{cases} \sqrt{-x+1}&\text{for }x < 1 \\\\ \sqrt{2x}&\text{for }x\geq 1 \end{cases}$ Is $h$ continuous at $x=1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
Explanation: For $h$ to be continuous at $x=1$, we need $\lim_{x\to 1}h(x)$ and $h(1)$ to exist and be equal. Since $1\geq 1$, the rule that applies to $x=1$ is $\sqrt{2x}$. So $h(1)=\sqrt{2(1)}=\sqrt{2}$. Now let's analyze $\lim_{x\to 1}h(x)$. Finding $\lim_{x\to 1^{ +}}h(x)$ For $x$ -values larger than $1$, the appropriate rule for $h(x)$ is $\sqrt{2x}$. Since $\sqrt{2x}$ is continuous for $x\geq1$, any limit in this interval is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to 1^{ +}}h(x) \\\\ &=\lim_{x\to 1^{ +}}[\sqrt{2x}] \gray{\sqrt{2x}\text{ is the rule for }x>1} \\\\ &=\sqrt{2(1)} \gray{\sqrt{2x}\text{ is continuous at }x=1} \\\\ &=\sqrt{2} \end{aligned}$ Finding $\lim_{x\to 1^{ -}}h(x)$ For $x$ -values smaller than $1$, the appropriate rule for $h(x)$ is $\sqrt{-x+1}$. Since $\sqrt{-x+1}$ is continuous for $x<1$, any limit in this interval is equal to the expression's value at that limit: $\begin{aligned} &\phantom{=}\lim_{x\to 1^{ -}}h(x) \\\\ &=\lim_{x\to 1^{ -}}[\sqrt{-x+1}] \gray{\sqrt{-x+1}\text{ is the rule for }x<1} \\\\ &=\sqrt{-1+1} \gray{\sqrt{-x+1}\text{ is continuous at }x=1} \\\\ &=0 \end{aligned}$ Conclusion We found that $\lim_{x\to 1^{ +}}h(x)=\sqrt{2}$ and $\lim_{x\to 1^{ -}}h(x)=0$. Since the one-sided limits aren't equal, $\lim_{x\to 1}h(x)$ doesn't exist and $h$ isn't continuous at $x=1$.